Question: Simplify the following expression: $y = \dfrac{3x^2+23x+40}{x + 5}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(3)}{(40)} &=& 120 \\ {a} + {b} &=& &=& {23} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $120$ and add them together. The factors that add up to ${23}$ will be your ${a}$ and ${b}$ When ${a}$ is ${8}$ and ${b}$ is ${15}$ $ \begin{eqnarray} {ab} &=& ({8})({15}) &=& 120 \\ {a} + {b} &=& {8} + {15} &=& 23 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({3}x^2 +{8}x) + ({15}x +{40}) $ Factor out the common factors: $ x(3x + 8) + 5(3x + 8)$ Now factor out $(3x + 8)$ $ (3x + 8)(x + 5)$ The original expression can therefore be written: $ \dfrac{(3x + 8)(x + 5)}{x + 5}$ We are dividing by $x + 5$ , so $x + 5 \neq 0$ Therefore, $x \neq -5$ This leaves us with $3x + 8; x \neq -5$.